IGCSE Maths May 2013 – Paper 1F Mark scheme (Final draft...

Mark Scheme (Results) January 2017 International GCSE Mathematics A 4MA0/3H

www.dynamicpapers.com

Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the world’s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information, please visit our website at www.edexcel.com. Our website subject pages hold useful resources, support material and live feeds from our subject advisors giving you access to a portal of information. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Expert email service helpful. www.edexcel.com/contactus Pearson: helping people progress, everywhere Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We’ve been involved in education for over 150 years, and by working across 70 countries, in 100 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: www.pearson.com/uk Janaury 2017 Publications Code 4MA0_3H_1701_MS All the material in this publication is copyright © Pearson Education Ltd 2017


http://www.edexcel.com/

http://www.edexcel.com/contactus

http://www.pearson.com/uk

General Marking Guidance • All candidates must receive the same treatment. Examiners

must mark the first candidate in exactly the same way as they mark the last.

• Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions.

• Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie.

• There is no ceiling on achievement. All marks on the mark scheme should be used appropriately.

• All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate’s response is not worthy of credit according to the mark scheme.

• Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited.

• When examiners are in doubt regarding the application of the mark scheme to a candidate’s response, the team leader must be consulted.

• Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

• Types of mark o M marks: method marks o A marks: accuracy marks o B marks: unconditional accuracy marks (independent of

M marks) • Abbreviations

o cao – correct answer only o ft – follow through o isw – ignore subsequent working o SC - special case o oe – or equivalent (and appropriate) o dep – dependent o indep – independent o eeoo – each error or omission


• No working If no working is shown then correct answers normally score full marks If no working is shown then incorrect (even though nearly correct) answers score no marks.

• With working If there is a wrong answer indicated on the answer line always check the working in the body of the script (and on any diagrams), and award any marks appropriate from the mark scheme. If it is clear from the working that the “correct” answer has been obtained from incorrect working, award 0 marks. Any case of suspected misread loses A (and B) marks on that part, but can gain the M marks. If working is crossed out and still legible, then it should be given any appropriate marks, as long as it has not been replaced by alternative work. If there is a choice of methods shown, then no marks should be awarded, unless the answer on the answer line makes clear the method that has been used. If there is no answer on the answer line then check the working for an obvious answer.

• Ignoring subsequent work It is appropriate to ignore subsequent work when the additional work does not change the answer in a way that is inappropriate for the question: eg. Incorrect cancelling of a fraction that would otherwise be correct. It is not appropriate to ignore subsequent work when the additional work essentially makes the answer incorrect eg algebra. Transcription errors occur when candidates present a correct answer in working, and write it incorrectly on the answer line; mark the correct answer.

• Parts of questions Unless allowed by the mark scheme, the marks allocated to one part of the question CANNOT be awarded in another.


International GCSE Maths Apart from questions 3, 5d, 17, 18, 20 & 23 (where the mark scheme states otherwise) the correct answer, unless clearly obtained from an incorrect method, should be taken to imply a correct method.

Q Working Answer Mark Notes 1 (a) 18 ÷ 60 oe or

7.3 or 18760

or 3710

or 7 × 60 + 18 (=438)

3 M1 for changing time to a decimal (7.3)

750 × “7.3” oe or 750 × "438"60

oe M1 for speed × time

(allow 750 × 7.18 or answer of 5385) 5475 A1

(b) for at least one correct operation eg. 750 × 1000, 750 ÷ 60 or

1000 ( 0.27....)60 60

=×

or 518

3 M1 for one or two of ×1000, ÷60, ÷60 (can be implied by 750 000 or 12.5 or12500 or 0.2083)

750 100060 60××

oe M1

complete correct method

208 A1 accept answers in range 208 – 208.3

•

Alternative mark scheme ft from (a)

“5475” × 1000 (=5475000) OR 7 × 60 + 18 = 438 and 438 × 60 (=26280 (sec))

3 M1

“5475000” ÷ 26280 M1 dep complete correct method

208 A1 accept answers in range 208 – 208.3

•

Total 6 marks


2 3 × 7 (=21) 2 M1

or for 3 numbers with a total of 21 or 3 numbers with a median of 5 or 3 numbers with a range of 14 or (a + c =) 3 × 7 – 5 (=16)

1, 5, 15 A1 numbers can be in any order Total 2 marks


3 173

− 195

3 M1

for correct improper fractions (subtraction sign not necessary) OR two improper fractions with a common denominator with at least one of the fractions correct

E.g. 8515

− 5715

or

17 5 3 1915

× − × oe

M1

for correct fractions with a common denominator a multiple of 15 i.e. in form 85𝑎

15𝑎 − 57𝑎15𝑎

shown A1 dep on M2 for correct conclusion to

15131 from correct working with sight of

the result of the subtraction e.g.1528

Alternative method (5)10

15 – (3) 12

15 3 M1

for two correct fractions with a common denominator a multiple of 15

− 215

M1

shown A1 dep on M2 for correct conclusion to

15131 from correct working with sight of

the result of the subtraction e.g.1528 or 2 −

215

Alternative method

E.g. 51015

– 3 1215

3 M1 for two correct fractions with a common denominator a multiple of 15

E.g. 42515

– 3 1215

M1

for a complete correct method

shown

A1 dep on M2 for correct conclusion to

15131 from correct working

Total 3 marks


4 π × ( 70 – 2 × 15) or π × 40 (=125(.6…))

4 M1

oe

4 × 15 (=60) and 4 × 70 (=280) or 340

M1

independent

“125.6...” + “60” + “280”

M1

dep on M2

466 A1 for answer in range 465.6 – 466

Total 4 marks


5 (a) h (7 + h) 1 B1 (b) 4p + 20 + 7p −14

2 M1

Any 3 terms correct

11p + 6 A1 cao NB 11p + 6 followed by, for example, 17p scores M1 A0

(c) 7 × (−2)² + 5 or 7 × 4 + 5 or 7 (−2)² + 5

2 M1

for correct substitution or 7 × 4 or 28

33 A1 (d) 5q – 15 (= 12 – q)

or 55

123 qq −=−

3 M1

E.g. 5q + q = 12 + 15 or 6q = 27

M1

For a correct equation with the q terms collected on one side of the equation and the non q terms on the other side. ft from 5q – 3 = 12 – q for this mark only

4.5 A1 for 4.5 or 9

2oe dep on at least M1

(e) −7t ≥ 31 – 3 or 7t ≤ 3 – 31 oe

2 M1

−7t ≥ 31 – 3 or 7t ≤ 3 – 31 or − 4 or t ≥ −4 accept an equation or the wrong inequality sign in the working

t ≤ −4 A1 or for −4 ≥ t Total 10 marks


6 2.5 × 28 + 7.5 × 32 + 12.5 × 20 + 17.5 × 14 + 22.5 × 6 or 70 + 240 + 250 + 245 + 135 or 940

4 M2

f × d for at least 4 products with correct mid- interval values and intention to add. If not M2 then award M1 for d used consistently for at least 4 products within interval (including end points) and intention to add or for at least 4 correct products with correct mid-interval values with no intention to add

(2.5 × 28 + 7.5 × 32 + 12.5 × 20 + 17.5 × 14 + 22.5 × 6) ÷ 100 or (70 + 240 + 250 + 245 + 135) ÷ 100 or “940” ÷ 100

M1

dep on M1 NB: accept their 100 if addition shown

9.4 A1

SC: B2 for answer of 9.44 (B1 for 944 in working)

Total 4 marks

7 96 ÷ 3 (= 32)

3 M1

M2 for 5 96

3×

9 × ‘32’(=288) or 4 × ‘32’(=128) or (9 − 4) × ‘32’

M1

dep

160 A1 Total 3 marks


8 (a) (−1, 6) (0, 4) (1, 2) (2, 0) (3, −2) (4, −4) (5, −6)

Correct line between x = −1 and x = 5

4 B4

For a correct line between x = −1 and x = 5

B3

For a correct line through at least 3 of (−1, 6) (0, 4) (1, 2) (2, 0) (3, −2) (4, −4) (5, −6) OR for all of (−1, 6) (0, 4) (1, 2) (2, 0) (3, −2) (4, −4) (5, −6) plotted but not joined.

B2

For at least 2 correct points plotted

B1 For at least 2 correct points stated (may be in a table) or seen in working OR for a line drawn with a negative gradient through (0, 4) OR for a line with the correct gradient.

(b)

3 M1

for y = −4 drawn; accept full or dashed line NB A shaded rectangle implies a choice of lines so M0

M1

for x = 1 drawn; accept full or dashed line NB A shaded rectangle implies a choice of lines so M0

For correct region identified

A1ft for correct region identified. Condone no label if region clear. ft from an incorrect straight line in part (a)

Total 7 marks


9 4x² + 6x + 6x + 9 or 4x² + 12x + 9

3 M1

for at least 3 terms correct in expansion of first pair of brackets

2x² − 10x + 3x – 15 or 2x2 – 7x – 15

M1 for at least 3 terms correct in expansion of second pair of brackets or all 4 terms correct ignoring signs allow –2x2 – 7x – 15

2x² + 19x + 24 A1

Alternative method (2x + 3)[(2x + 3) – (x – 5)]

M1

(2x + 3)(x + 8)

M1

2x² + 19x + 24 A1 Total 3 marks

10 0.82x = 25.83 or 82% = 25.83

3 M1

or for use of 0.82 in a calculation

25.830.82

or 25.8382

× 100

M1

31.5(0) A1 Total 3 marks


11 (a) 4, 16, 42, 84, 96, 100 4, 16, 42, 84, 96, 100 1 B1 cao (b) (20, 4) (40, 16) (60, 42) (80, 84)

(100, 96) (120, 100)

2 M1 (ft from sensible table i.e. clear attempt at addition) for at least 4 points plotted correctly at end of interval or for all 6 points plotted consistently within each interval in the freq table at the correct height

correct cf graph A1 accept curve or line segments accept curve that is not joined to (0,0)

(c) 46 - 48 1 B1 ft from a cumulative frequency graph (d) E.g. reading from graph at t = 70

2 M1 for evidence of using graph at t = 70

ft from a cumulative frequency graph provided method is shown

36 – 38 A1 100 – ‘63’ ft from a cf graph ft from a cumulative frequency graph provided method is shown

Total 6 marks


12 (a)(i) 2 × 48 96 1 B1 (ii)

The angle at the centre is double the

angle at the circumference

1 B1 NB : accept twice, double, origin (O) accept ‘angle at circumference is half the angle at the centre’ oe

(b) (i)

180 − 48 132 1 B1

(ii) The opposite angles in a cyclic

quadrilateral total 180°

1 B1 accept supplementary angles accept The angle at the centre is double the angle at the circumference with angles at a point sum to 360o

Total 4 marks

13 0.0275 × 4000 (=110)

3 M1

for interest for first year or 330 or answer of 4330

M2 for 1.02753 × 4000 oe

E.g. 0.0275 × (4000 + “110”) (=113.025) and 0.0275 × (4000 + “110” + “113.025”)

M1

for a complete method

4339.16 A1 Accept answer in range 4339 – 4340 NB: Answer in range 339 – 340 gets M2A0

Total 3 marks


14 (a) T = k√𝑥

3 M1

or for T mx= k may be numeric (but not 1)

400 = k√625 or k = 16 or 400 625m= or m = 256

M1

implies the first M1

T =16√𝑥 A1 accept 256T x= Award 3 marks if T = k√𝑥 but k is evaluated correctly in part (a) or (b). SC: B2 for correct formula for x in terms of T

(b) 120 1 B1 ft for a correct answer from a substitution into an equation (or expression) in the form (T =) k√𝑥 except for k = 1

Total 4 marks


15 (x² =) 17² + 14² − 2 × 17 × 14 × cos(123°)

4 M1

E.g. (x² =) 744(.248......) or (x² =) 17² + 14² − −259(.2… )

M1

for correct order of operations

(x =)27.28.....

A1

for missing side in range 27.2 – 27.3

58.3 B1ft dep on M1 ft for “27.28” + 31 Alternative scheme (height =) 14 × sin(180 – 123) (=11.7…) M1

14 × cos(180 – 123) (=7.6…) M1

2 2"11.7" "(17 7.6)"+ + (=27.28) A1

58.3 B1ft dep on M1 ft for “27.28” + 31 Total 4 marks


16 (a) 3

264891

or 3891264

or 23

or or 32

oe or 2 : 3

or 3 3264 : 891 ( 6.415 : 9.622)=

2 M1

correct linear scale factor or correct ratio (numbers may be in either order)

12 A1 cao (b)

459 × �23�² oe or 459 ÷ �3

2�² oe or

459 × 41(.153…) ÷ 92(.594…)

2 M1

correct method to find the surface area of A

204 A1 cao Total 4 marks


17 −8 ± √52410

or

−8 ± √82− −460

2×5 oe or

−8 ± 2√131

10

NB: denominator must be 2×5 or 10 and there must be evidence for correct order of operations in the numerator

3 M2 If not M2 then M1 for

−8 ± √82 − 4 × 5 × −232 × 5

condone one sign error in substitution; allow partial correct evaluation

1.49, −3.09 A1 for answers in range 1.489 to 1.489105 and −3.089 to −3.0891045 Award M2 A1 for answers in range 1.489 to 1.489105 and −3.089 to −3.0891045 with sufficient correct working that would gain at least M1

Alternative scheme

5[(x + 45)² − 16

25] oe

3 M1 for completing the square

4 23 165 5 25

− ± + oe M1

1.49, −3.09 A1 for answers in range 1.489 to 1.489105 and −3.089 to −3.0891045

Total 3 marks


18 d 20 9dy xx

= +

M1

for differentiating 10x2 or 9x correctly M2 for 9

2 10−×

(from 2

ba− )

20x + 9 = 0

M1 equating their d

dyx

(of the form ax + b)

to zero, dep on previous M1 x = −0.45 oe A1

dep on at least M1 for x = −0.45 oe

(−0.45, 2.975) oe 4 A1ft dep on M2 accept fractions

9 39,220 40

−

or 9 119,20 40

−

Alternative scheme (completing the square)

29( )20

x + + ….. M1

229 9 5( ) 0

20 20 10x + − + =

M1

x = −0.45 oe A1

dep on at least M1 for x = −0.45 oe

(−0.45, 2.975) oe 4 A1ft dep on M2 accept fractions

9 39,220 40

−

or 9 119,20 40

−


19 2 5 2

3m ek

e+

= or 3 5 2k e m e= +

4 M1

Squaring both sides or clearing fraction

3ek² = 5m + 2e

M1

Clearing fraction and squaring both sides

3ek² − 2e = 5m or −5m = 2e – 3ek² e(3k² − 2) = 5m or −5m = e(2 – 3k²)

M1

Isolating terms in e in a correct equation

e = 5𝑚3𝑘2−2

A1

for 2

53 2

mek

=−

or 2

52 3

mek

−=

− oe

Total 4 marks

20 3.5 or 2.5 or 5.25 or 5.35 or 8.365 or 8.375

3 M1

accept 3.49̇ or 3.499… or 5.349 ̇ or 5.3499... or 8.3749̇ or 8.37499..

3.5(8.375 – 5.25) or 3.5 × 8.375 – 3.5 × 5.25

M1

or for UB1 ×(UB2 – LB) oe where 3 < UB1 ≤ 3.5 and 8.37 < UB2 ≤ 8.375 and 5.25 ≤ LB < 5.3

17516

or 10.9375 A1 dep on M2 – correct working must be seen

Total 3 marks


21 (a) 12 1 B1 (b) 7 1 B1 (c)

Correct region shaded

1 B1 Must be unambiguous

Total 3 marks

22 75360

× 𝜋 ×d (2r) = 7.2 4 M1 for a correct equation linking the angle and arc length

NB: 0.208(3… ) may be used in place of36075

or 4.8 in place of 36075

7.2 36075 π××

oe or 7.2 36075 2 π

×× ×

oe or

d = 11(.0…) or r = 5.5(0…)

M1 for a complete method to find the radius or diameter.

275 "11"360 2

π × ×

or 275 "5.5"360

π× × or

75 95(.04...)360

×

M1 dep on previous M1

19.8 A1 for answer in range 19.8 – 19.82 Total 4 marks


23 x² + (8−2x)² = 52 6 M1 for elimination of one variable

�8 −𝑦2� ² + y² = 52

x² + 4x² − 16x – 16x + 64 =52 M1 (indep) for a correct expansion of

(8 – 2x)² or �8 −𝑦2�²

�64−8𝑦−8𝑦+𝑦²4

� + y² = 52

5x² − 32x + 12 (= 0) A1 for correct simplified 3 term quadratic equation in any form (may not be equated to zero) 5y² − 16y – 144 (= 0)

(5x – 2)(x – 6) (=0)

or −−32 ±�(−32)² −4×5×12

2×5

(may be partially evaluated, condone lack of brackets around negative numbers)

M1 (5y – 36)(y + 4) (=0)

or −−16 ±�(−16)²−4×5×−144

2×5

(may be partially evaluated, condone lack of brackets around negative numbers) NB: can ft for this mark only provided M1 awarded and a 3 term quadratic

x = 25 or x = 6 A1 for both x values (or both y values)

y = 365

or y = −4 x = 2

5 oe, y = 36

5 oe

x = 6 , y = −4

A1 for both solutions with x and y values correctly paired

Total 6 marks


24 69 × 7

10 × 7

10 49

150 =

oe or 39 × 4

10 × 4

10 4

75 =

oe OR

69 × 7

10 × a and 3

9 × 4

10 × b

a and b must both be a single fraction where

0 < a, b < 1 and 7 4,10 10

a b≠ ≠

3 M1

69 × 7

10 × 7

10 oe and 3

9 × 4

10 × 4

10 oe

M1

Both products correct (addition not needed)

1950

A1

oe E.g. 342900

Total 3 marks


Pearson Education Limited. Registered company number 872828 with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom


IGCSE Maths May 2013 – Paper 1F Mark scheme (Final draft...

Documents

Transcript of IGCSE Maths May 2013 – Paper 1F Mark scheme (Final draft...